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Question Number 58675 by ajfour last updated on 27/Apr/19

Commented by ajfour last updated on 27/Apr/19

$ABCD is a square; while, PQM is$ $a sector . Find in what proportion$ $are radii b, r, g (blue, red, green$ $respectively) .$
Commented by ajfour last updated on 27/Apr/19

Commented by ajfour last updated on 27/Apr/19

$let square side be a .$ $\Rightarrow \frac{r}{\sin α} + r + 2 g = a$ $(g + b) \cos β + b = g$ $\frac{r}{\sin α} + r + g = \frac{g}{\cos α}$ $. . . . . . . .$ $. . . . . . . .$
Commented by mr W last updated on 01/May/19

$M P = M Q = M N = P Q = a$ $\Rightarrow Δ M P Q = e q u i l a t e r a l$ $\Rightarrow α = 60 °$ $E M = a - g$ $E H = g$ $E M = 2 E H$ $a - g = 2 g$ $\Rightarrow g = \frac{a}{3}$ $F G = r$ $F M = a - 2 g - r = 2 F G = 2 r$ $\Rightarrow r = \frac{a - 2 g}{3}$ $\Rightarrow r = \frac{a}{9}$ $M L = a - b$ $M J = \sqrt{{M L}^{2} - {J L}^{2}} = \sqrt{{(a - b)}^{2} - b^{2}} = \sqrt{a (a - 2 b)}$ $M H = \sqrt{3} E H = \sqrt{3} g = \frac{\sqrt{3} a}{3}$ $H J = M J - M H = \sqrt{a (a - 2 b)} - \frac{\sqrt{3} a}{3}$ ${H J}^{2} = {(g + b)}^{2} - {(g - b)}^{2} = 4 g b = \frac{4 a b}{3}$ $\Rightarrow a (a - 2 b) + \frac{a^{2}}{3} - \frac{2 a}{3} \sqrt{3 a (a - 2 b)} = \frac{4 a b}{3}$ $\Rightarrow \frac{4 a^{2}}{3} - \frac{2 a}{3} \sqrt{3 a (a - 2 b)} = \frac{10 a b}{3}$ $\Rightarrow 2 a - \sqrt{3 a (a - 2 b)} = 5 b$ $\Rightarrow 2 a - 5 b = \sqrt{3 a (a - 2 b)}$ $\Rightarrow 4 a^{2} + 25 b^{2} - 20 a b = 3 a^{2} - 6 a b$ $\Rightarrow 25 b^{2} - 14 a b + a^{2} = 0$ $\Rightarrow b = \frac{7 - 2 \sqrt{6}}{25} a \approx \frac{a}{11.9} = 0.084 a < r$ $\Rightarrow g / r / b = 3 / 1 / 0.756$
Commented by ajfour last updated on 01/May/19

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